Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance MCQs with answers

Answer: (a) 300 V

V = (1/4πε₀) q/r = 9 × 10⁹ × 5.0 × 10⁻⁹ / 0.15 = 45/0.15 = 300 V. Dividing by r² gives 2000 V (the field's form, not the potential's); leaving out r gives 45 V; taking r as 1.5 cm gives 3000 V.

Answer: (b) zero and −V₀

V = p cos θ/(4πε₀r²). At (0, 20 cm, 0), θ = 90° (equatorial plane), so V = 0. At (−20 cm, 0, 0), on the axis at the same distance, θ = 180°, so V = −V₀: the point is nearer −q.

Answer: (c) 1000 eV

Energy gained = qΔV = 2e × 500 V = 1000 eV. 500 eV is the value for a charge e, as for a proton or electron.

Answer: (d) concentric spheres centred at the charge

V = q/(4πε₀r) depends only on r, so V is constant wherever r is constant: on spheres centred at the charge. The radial field lines cross them normally.

Answer: (d) Q/(4πε₀R)

The field inside the shell is zero, so no work is done moving a test charge inside it: the potential inside is constant and equals the surface value Q/(4πε₀R). Zero is the field there, not the potential.

Answer: (a) 400 pF

C = ε₀A/d = 8.85 × 10⁻¹² × 4.0 × 10⁻² / (0.885 × 10⁻³) = 4.0 × 10⁻¹⁰ F = 400 pF.

Answer: (d) 2 μF

1/C = 1/6 + 1/3 = 1/2, so C = 2 μF. 9 μF is the parallel value; in series the combination is smaller than the smallest capacitor.

Answer: (b) 60 μC

In parallel each capacitor has the full 12 V across it: Q = CV = 5 μF × 12 V = 60 μC. 84 μC is the total charge; 24 μC is on the 2 μF capacitor.

Answer: (b) 0.40 J

U = ½QV = ½ × 4.0 × 10⁻³ × 200 = 0.40 J. QV alone gives 0.80 J; halving twice gives 0.20 J.

Answer: (d) [M⁻¹ L⁻² T⁴ A²]

C = Q/V: charge is [A T] and potential is [M L² T⁻³ A⁻¹], so C = [A T]/[M L² T⁻³ A⁻¹] = [M⁻¹ L⁻² T⁴ A²]. [M L² T⁻³ A⁻¹] is the dimension of potential itself.